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Let $\mathcal{H}_n \mathbb{C}$ be the set of hermitian $n \times n$ complex matrices. This set carries the structure of a vector space over $\mathbb{R}$ under usual addition. It also inherits the standard euclidean topology from $\mathbb{C}^{n\times n}$. Let $\mathcal{P}$ denote the subset of $\mathcal{H}_n \mathbb{C}$ of positive definite matrices and give it the subspace topology.

My question is: is $\mathcal{P}$ locally compact?

What I understand: $\mathcal{P}$ is closed under linear combination with positive coefficients, in particular it is closed under addition, multiplication by a positive scalar and is convex. In dimension $1$ it is just $]0,+\infty[$, so the answer is yes. My feeling is that the answer should be yes in higher dimensions as well, since $\mathcal{P}$ is a kind of "cone" in a real vector space, but I can't provide a rigorous proof of this.

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I think this reasoning would apply to every subset of a metric space, but $\mathbb{R}$ has already some subset which is not locally compact, e.g. $\mathbb{Q}$. The point is that the intersection of the unit ball with the subspace need not be compact. –  Lor Apr 6 '13 at 10:19
    
But my question was about $\mathcal{P}$, which is not. See my comment to your answer below. –  Lor Apr 6 '13 at 15:49

2 Answers 2

up vote 2 down vote accepted

Edit: I think that the first answer below is more efficient and easier to write. Nevertheless, I have added a more concrete approach which might be closer to what you were after.

By some sort of transitivity: recall that every open or closed subset of a locally compact space is a locally compact space with the induced topology. We will use both, for open and for closed.

I consider every space here equipped with the topology induced by the Euclidean norm of $\mathbb{C}^{n\times n}$. Hence every space is Hausdorff. This is good to know, even though your question does not specifically ask about this aspect.

Like every finite-dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$, $\mathbb{C}^{n\times n}$ is locally compact when equipped with the topology induced by any norm.

Clearly, $\mathcal{H}_n^+\mathbb{C}$, the set of positive semidefinite matrices, is closed in the locally compact $\mathbb{C}^{n\times n}$. So $\mathcal{H}_n^+\mathbb{C}$ is a locally compact space.

Now $\mathcal{P}=\{A\in \mathcal{H}_n^+\mathbb{C}\;;\det A>0\}$. By continuity of the determinant, it follows that $\mathcal P$ is open in the locally compact $\mathcal{H}_n^+\mathbb{C}$. Hence $\mathcal P$ is a locally compact space. QED.

Parametrized alternative: fix $A_0$ hermitian definite positive, and denote $\{t^0_1,\ldots,t_n^0\}$ its (positive) eigenvalues. Now let $\epsilon:=\min t_j^0/2>0$. Then denote $U_n$ the unitary group and $\mathcal{H}_n^{++}$ the cone of positive definte hermitian matrices. Now consider the map $$ \phi:U_n\times \prod_{j=1}^n[t_j^0-\epsilon,t_j^0+\epsilon]\longrightarrow \mathcal{H}_n^{++} $$ which sends $(U,t_1,\ldots,t_n)$ to $U\mbox{diagonal}\{t_1,\ldots,t_n\}U^*$. Since $U_n$ is compact, the domain is compact. And since $\phi$ is continuous, the range of $\phi$ is a compact subset of $\mathcal{H}_n^{++}$ containing $A_0$. So it only remains to check that this is a neighborhood of $A_0$ in $\mathcal{H}_n^{++}$. To that aim, note that it contains $$ \phi(U_n\times \prod_{j=1}^n(t_j^0-\epsilon,t_j^0+\epsilon))\ni A_0 $$ i.e. the set of hermitian definite positive matrices with spectrum $\{t_1,\ldots,t_n\}$ such that, up to a permutation, $|t_j-t_j^0|<\epsilon$ for every $j=1,\ldots,n$. By continuity of polynomial roots over $\mathbb{C}$ applied to the characteristic polynomial, this is open in $\mathcal{H}_n^{++}$. QED.

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@Lor Do you agree with my argument? –  1015 Apr 6 '13 at 20:02
    
Yes, thank you. I think this works! –  Lor Apr 7 '13 at 15:52
    
You're welcome. I think it works too. Isn't it funny we have to do it in two somehow symmetric (closed/open) steps? –  1015 Apr 7 '13 at 15:55
    
Yes it is. In dimension $1$ we have $]0,+\infty[ \subset \mathbb{R} \subset \mathbb{C} $, so we can see already from here that there's some non trivial subspace topology going on. I was hoping for a proof involving the convexity property of $\mathcal{P}$, but I'll be satisfied with yours. –  Lor Apr 7 '13 at 16:04
    
@Lor I'v added another approach, which might be closer to what you were hoping for. –  1015 Apr 7 '13 at 16:58

Edit: First, as $\mathbb{C}^{n\times n}$ is Hausdorff, so is $\mathcal{P}$.

Second, let $B(A,r)$ denotes the ball centred at a matrix $A$ with radius $r$ in $\mathbb{C}^{n\times n}$ and $\bar{B}(A,r)$ its closure. Since eigenvalues are continuous functions of matrix entries, for any $P\in\mathcal{P}$, there exists $r>0$ such that $B(P,r)\cap(\mathcal{H}_n \mathbb{C})\subset\mathcal{P}$. Therefore, for all $\varepsilon\in(0,r)$, the closure of $B(P,\varepsilon)\cap\mathcal{P}$ in $\mathcal{P}$ is simply $\bar{B}(P,\varepsilon)\cap\mathcal{P}=\bar{B}(P,\varepsilon)\cap(\mathcal{H}_n \mathbb{C})$, which is closed and bounded and hence compact in $\mathbb{C}^{n\times n}$. As compactness is an intrinsic property of a set, $\bar{B}(P,\varepsilon)\cap\mathcal{P}$ is also compact in $\mathcal{P}$. Furthermore, $\bar{B}(P,\varepsilon)\cap\mathcal{P}$ has a non-empty interior ($B(P,\varepsilon)\cap\mathcal{P}$). So, it is a compact neighbourhood in $\mathcal{P}$.

In other words, $\mathcal{P}$ is a Hausdorff space such that its every member is contained in a compact neighbourhood. Hence $\mathcal{P}$ is locally compact.

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I'm fine with this. After all it's a real vector space. But my question was about the topological subspace of positive definite hermitian matrices. This is not a vector subspace, nor a closed subspace, hence my doubts. –  Lor Apr 6 '13 at 15:48
    
@Lor Thanks. I am talking about topological subspace, not vector subspace. (Why would you think I was talking about vector subspace?) But I have overlooked the keyword "positive definite", so I thought you only meant Hermitian matrices. I'll see if the answer can be fixed. If not, I'll delete this answer. –  user1551 Apr 6 '13 at 16:45
    
This is very straightforward, thanks! But I think we have to use that $\mathcal{P}$ is an open subspace in a closed one as @julien does in the answer above. Otherwise we can't guarantee that a subspace which happens to be compact in the ambient topology will also be compact in the subspace topology. Am I wrong? I don't want to be too pedantic but I think this is the key point. –  Lor Apr 7 '13 at 15:45
    
I mean that $B(P,\varepsilon) \cap \mathcal{P}$ is of course an open neighborhood of $P$ in $\mathcal{P}$ and is contained in $\bar{B}(P,\varepsilon) \cap \mathcal{P}$ which is compact in $C^{n\times n}$. But I can't conclude from here that it is also compact in $\mathcal{P}$ if I don't know something more about the topology of $\mathcal{P}$. –  Lor Apr 7 '13 at 15:49
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@Lor A set that is compact in a superspace is also compact in a subspace. (In fact, compactness is an intrinsic property of a set. The compactness of a set depends only on the topology of itself. You may speak of the compactness of a set without making any reference to its proper superspace.) –  user1551 Apr 8 '13 at 8:49

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